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3 years ago

Discuss the phase change condition due to reflection of light from a surface. Summarize equations of interference for thin film.

Physics

1 answer:

Dmitrij [34]3 years ago

5 0

Answer:

if this surface has a higher index than in the medium where the light travels, the reflected wave has a phase change of 180º

Explanation:

When a ray of light falls on a surface if this surface has a higher index than in the medium where the light travels, the reflected wave has a phase change of 180º this can be explained by Newton's third law, the light when arriving pushes the atoms of the medium that is more dense, and these atoms respond with a force of equal magnitude, but in the opposite direction.

When the fractional index is lower than that of the medium where the reflacted beam travels, notice a change in phase.

Also, when light penetrates the medium, it modifies its wavelength

λ = λ₀ / n

We take these two aspects into account, the condition for contributory interference is

d sin θ = (m + 1/2) λ

for destructive interference we have

d sin θ = m λ

in general this phenomenon is observed at 90º

2 d = (m +1/2) λ° / n

2nd = (m + ½) λ₀

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Compare the energy consumption of two commonly used items in the household. Calculate the energy used by a 1.40 kW toaster oven,

andrew-mc [135]

Energy = (power) x (time)

-- <u>For the toaster:</u>

Power = 1.4 kW = 1,400 watts

Time = 5.4 minutes = 324 seconds

Energy = (1,400 W) x (324 s) = 453,600 Joules

-- <u>For the CFL bulb:</u>

Power = 11 watts

Time = 10.5 hours = 37,800 seconds

Energy = (11 W) x (37,800 s) = 415,800 Joules

-- The toaster uses energy at 127 times the rate of the CFL bulb.

-- The CFL bulb uses energy at 0.0079 times the rate of the toaster.

-- The toaster is used for 0.0086 times as long as the CFL bulb.

-- The CFL bulb is used for 116.7 times as long as the toaster.

-- The toaster uses 9.1% more energy than the CFL bulb.

-- The CFL bulb uses 8.3% less energy than the toaster.

7 0

2 years ago

What tripositive ion has the electron configuration [kr] 4d3 ? what neutral atom has the electron configuration [kr] 5s 2 4d2 ?

tigry1 [53]

(a)

Electronic configuration is given as follows:

$[Kr]4d^{3}$

Since, this is the electronic configuration of ion with+3 that means 3 electrons are removed. On adding the 3 electrons, the electronic configuration of neutral atom can be obtained.

Thus, electronic configuration of neutral atom is $[Kr]4d^{5}5s^{1}$ .

The atomic number of Kr is 36, thus, total number of electrons become 36+6=42.

This corresponds to element: molybdenum. Thus, the tripositive atom will be $Mo^{3+}$ .

(b) The given electronic configuration is $[Kr]5s^{2}4d^{2}$ .

The atomic number of Kr is 36, thus, total number of electrons become 36+4=40.

This corresponds to element zirconium, represented by symbol Zr.

6 0

3 years ago

g Two masses are involved in a collision on an axis (one dimensional). One mass is six times the mass of the second. Both masses

statuscvo [17]

Answer:

v₁f = 0.5714 m/s (→)

v₂f = 2.5714 m/s (→)

e = 1

It was a perfectly elastic collision.

Explanation:

m₁ = m

m₂ = 6m₁ = 6m

v₁i = 4 m/s

v₂i = 2 m/s

v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i + ((2m₂) / (m₁ + m₂)) v₂i

v₁f = ((m – 6m) / (m + 6m)) * (4) + ((2*6m) / (m + 6m)) * (2)

v₁f = 0.5714 m/s (→)

v₂f = ((2m₁) / (m₁ + m₂)) v₁i + ((m₂ – m₁) / (m₁ + m₂)) v₂i

v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)

v₂f = 2.5714 m/s (→)

e = - (v₁f - v₂f) / (v₁i - v₂i) ⇒ e = - (0.5714 - 2.5714) / (4 - 2) = 1

It was a perfectly elastic collision.

8 0

2 years ago

when an object moves with constant velocity, does its average velocity during any time interval differ from its instataneous vel

BigorU [14]

For a constant-velocity object, the average and instantaneous are the same. So the answer is no. It's like taking a running average of a string of numbers that are all the same number. The average is always the sum of the numbers divided by how many have accumulated, which will always equate to the repeated number.

8 0

3 years ago

You are designing a flywheel. It is to start from rest and then rotate with a constant angular acceleration of 0.200 rev/s^2. Th

Rama09 [41]

Answer:

I = 8.75 kg m

Explanation:

This is a rotational movement exercise, let's start with kinetic energy

K = ½ I w²

They tell us that K = 330 J, let's find the angular velocity with kinematics

w² = w₀² + 2 α θ

as part of rest w₀ = 0

w = √ 2α θ

let's reduce the revolutions to the SI system

θ = 30.0 rev (2π rad / 1 rev) = 60π rad

let's calculate the angular velocity

w = √(2 0.200 60π)

w = 8.683 rad / s

we clear from the first equation

I = 2K / w²

let's calculate

I = 2 330 / 8,683²

I = 8.75 kg m

4 0

2 years ago